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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 4 SOLUTIONS 1. Let Ω ⊆ C be a region. Let f = u + iv be analytic on Ω. (a) If αu + βv is constant on Ω for some α,β ∈ C × , show that f is constant on Ω. Solution. Taking partial derivatives of αu + βv = constant , we get ( αu x + βv x = 0 , αu y + βv y = 0 . Substituting the CauchyRiemann equations u x = v y and u y = v x gives ( αu x βv y = 0 , αu y + βu x = 0 . Taking complex conjugate of the second equation (and noting that u x and u y are real valued) and writing the resulting system in matrix form yields α β β α u x u y = . Since α,β 6 = 0, det α β β α =  α  2 +  β  2 6 = 0 , and thus the system has a unique solution u x = 0, u y = 0. Applying the CauchyRiemann equations again, we get v x = 0, v y = 0. Since Ω is a region, we must have that f is constantvalued on Ω. (b) If u 2 + v 2 is constant on Ω, show that f is constant on Ω. Solution. Let the constant be c . Then u 2 + v 2 = c. If c = 0, then u = 0 and v = 0 and so f = 0, thus constant. We will assume that c 6 = 0. Taking partial derivatives, we get ( uu x + vv x = 0 , uu y + vv y = 0 . (1.1) Substituting the CauchyRiemann equations u x = v y and u y = v x into the second equation of (5.12) gives ( uu x + vv x = 0 , uv x + vu x = 0 . (1.2) Now we multiply the first equation in (1.2) by u on both sides and use the second equation in (1.2) to get 0 = u ( uu x + vv x ) = u 2 u x + v ( uv x ) = u 2 u x + v ( vu x ) = ( u 2 + v 2 ) u x = cu x . Date : October 21, 2009 (Version 1.0). 1 Similarly, we could have instead substitute the CauchyRiemann equations u x = v y and u y = v x into the first equation of (5.12) to get ( uv y vu y = 0 , uu y + vv y = 0 . (1.3) This time, multiplying the second equation in (4.10) by v on both sides and use the first equation in (4.10) yields 0 = v ( uu y + vv y ) = u ( vu y ) + v 2 v y = u ( uv y ) + v 2 v y = ( u 2 + v 2 ) v y = cv y . Since c 6 = 0, we get u x = 0 and v y = 0 . Applying the CauchyRiemann equations again, we get v x = 0, u y = 0. Since Ω is a region, we must have that f is constantvalued on Ω. (c) If u = v 2 , show that f is constant on Ω. Solution. Applying the CauchyRiemann equations and chain rule, we get 2 vv x = v y , 2 vv y = v x . Eliminating v x from the equations, we get 4 v 2 v y = v y . Suppose v y 6≡ 0. Then there must exist a point z = x + iy ∈ Ω such that v y ( x ,y ) 6 = 0. So v ( x ,y ) 2 = 1 / 4 — which is impossible since v is realvalued. Hence we must have v y ≡ 0 and in which case, we apply the CauchyRieman equations again to get v x ≡ v y ≡ u x ≡ u y ≡ . Hence f is constant. (d) If u = ϕ ◦ v for some differentiable real function ϕ : R → R , show that f is constant on Ω....
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This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Lim
 Math, Derivative

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